Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 0-32198-238-X
ISBN 13: 978-0-32198-238-4

Chapter 1 - Linear Equations in Linear Algebra - 1.7 Exercises: 41

Answer

$\mathbf{B}=\begin{bmatrix}8&-3&2\\-9&4&-7\\6&-2&4\\5&-1&10\end{bmatrix} $

Work Step by Step

We wish to find a linearly indepent set of columns of $\mathbf{A}$. Using a CAS to find $\text{rref}(\mathbf{A})$, we discover that the third and fourth columns do not contain pivots. This tells us that the third column lies in the span of the first two columns and that the fourth column lies in the span of the first three. But, without calculating further, we can deduce that the fourth column is in the span of the first two, since the third column does not contribute to their span. Hence, we delete columns two and three and confirm that the $4\times 3$ matrix above yields {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}, {0, 0, 0}} when the command RowReduce[{{8, -3, 2},{-9, 4, -7},{6, -2, 4},{5, -1, 10}}] is entered into Mathematica. Results using Matlab and Maple should be similar.
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